[LeetCode] 387. First Unique Character in a String 字符串的第一个唯一字符

Given a string, find the first non-repeating character in it and return it‘s index. If it doesn‘t exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.

Note: You may assume the string contain only lowercase letters.

给一个字符串，找出第一个不重复的字符，返回它的index，如果不存在，返回-1。假设字符都是小写字母。

解法1: 暴力搜索， T：O(n^2)

解法2: HashMap，第一次遍历每一个字符，统计每种字符出现的次数。第二次遍历，找到第一出现的字符次数为1的字符。

解法3: int [26]数组，由于只有26个字母，所以可以用数组来统计出现的个数。

Java: Brute Force

class Solution {
    public static int firstUniqChar(String s){
        for (int i = 0; i < s.length(); i++) {
            boolean isUnique = true;
            for (int j = 0; j < s.length(); j++) {
                if (i != j && s.charAt(i) == s.charAt(j)){
                    isUnique = false;
                    break;
                }
            }
            if (isUnique) return i;
        }
        return -1;
    }
}　　

Java:

class Solution {
    public int firstUniqChar(String s){
        Map<Character, Integer> charMap = new HashMap<>(s.length()); //预先分配大小,避免扩容性能影响
        for (int i = 0; i < s.length(); i++) {
            if (!charMap.containsKey(s.charAt(i))){
                charMap.put(s.charAt(i), 1);
            }else {
                charMap.put(s.charAt(i), charMap.get(s.charAt(i))+1);
            }
        }

        for (int i = 0; i < s.length(); i++) {
            if (charMap.get(s.charAt(i)) == 1){
                return i;
            }
        }
        return -1;
    }
}　　

Java:

public class Solution {
    public int firstUniqChar(String s) {
        char[] array = s.toCharArray();
        int[] a = new int[26];
        for(int i=0;i<s.length();i++)a[array[i]-‘a‘]++;
        for(int i=0;i<s.length();i++){
            if(a[array[i]-‘a‘]==1){
                return i;
            }
        }
        return -1;
    }
}

Java:

class Solution {
    public int firstUniqChar(string s) {
        vector<int> count(26);
        for(int i=0;i<s.size();i++)
            count[s[i]-‘a‘]++;
        for(int i=0;i<s.size();i++)
            if(count[s[i]-‘a‘]==1)
                return i;
        return -1;
    }
}

Java:　　

class Solution {
    public int firstUniqChar(String s) {
      for(int i = 0; i<s.length(); i++) {
        if(s.lastIndexOf(s.charAt(i))==s.indexOf(s.charAt(i))) return i;
       }
       return -1;
    }
}　　

Python:

class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        letters = {}
        for c in s:
            if c in letters:
                letters[c] = letters[c] + 1
            else:
                letters[c] = 1
        for i in xrange(len(s)):
            if letters[s[i]] == 1:
                return i
        return -1　　

Python: 162ms

from collections import defaultdict

class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        lookup = defaultdict(int)
        candidtates = set()
        for i, c in enumerate(s):
            if lookup[c]:
                candidtates.discard(lookup[c])
            else:
                lookup[c] = i+1
                candidtates.add(i+1)

        return min(candidtates)-1 if candidtates else -1

Python: 92ms

class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        return min([s.find(c) for c in ‘abcdefghijklmnopqrstuvwxyz‘ if s.count(c)==1] or [-1])

Python: 75ms

class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        return min([s.find(c) for c in string.ascii_lowercase if s.count(c)==1] or [-1])　　

C++:

class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, int> m;
        for (char c : s) ++m[c];
        for (int i = 0; i < s.size(); ++i) {
            if (m[s[i]] == 1) return i;
        }
        return -1;
    }
};

原文地址：https://www.cnblogs.com/lightwindy/p/8655155.html